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Homework Help: Change of momentum and kinetic energy

  1. Jan 8, 2005 #1
    Im probably being stupid here but Im confused as to when you can use the impulse formula: I = F*t and when you use F*d = 0.5mv^2 for finding a constant applied force.

    If a particle's moving at a constant velocity decelerating to 0m/s in a set distance when acted upon by a constant force f, the force I calculate is double when I use I = F*t (momentum lost = force * time of application) than that of when I use
    KE lost = Work done = F*d

    e.g m= 3Kg u = 4m/s stopping distance(x)= 5m

    Using momentum: I = mv - mu = 12Ns

    meanwhile v=4 m/s
    therefore dx = 4t by integration
    ds = 5m
    therefore t = 5/4 s

    So using I = F*t
    F= I/t
    F= 12/1.25
    F= 9.6N

    Using K.E
    KE lost = Work done = F*x

    F*x = 0.5mv^2
    F*x = 0.5*3*4^2
    F*x = 24
    therefore F= 24/5
    F= 4.8 N

    Why is this half??? :confused:
  2. jcsd
  3. Jan 8, 2005 #2


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    So you're trying to compute the force...This means finding the acceleration,which means that the velocity is not constant,which means that your integration was wrong...The whole post is wrong,actually.

    For constant accelerated movement,we know Galilei's formula:
    [tex] v^{2}_{fin.}=v^{2}_{init.}+2a x [/tex]
    If the final velocity is 0,and the initial one is 4m/s,then,for the distance of movement x=5m,then the accleration is negative and found to be
    [tex] a=-\frac{8}{5}ms^{-2} [/tex]
    For a body of mass "m=3kg",the force of deceleration is negative as well and
    [tex] F=ma=-\frac{24}{5}N=-4.8 N [/tex]
    ,which is exactly what u get by applying the KE conservation law.
    U did that wrongly and u ended up without the minus.

  4. Jan 8, 2005 #3
    This wrong because you also need to calculate the negative acceleration that expresses the fact that you slow down. x =4t is wrong...

    You can easely calculate this velocity if you know that the initial velocity is 4 m/s and the stopping distance is 3 meters. Just use : x = x_0 + 4t -at²/2 and (x_0 is the initial position)

    v = 4 -at. When x-x_0 is 3 you have stopped thus v = 0. so t = 4/a and substitute this in the first equation to find the magnitude of a. Remeber that is a MUST be negative...

    then move on just like you did...
    Last edited: Jan 8, 2005
  5. Jan 9, 2005 #4

    Andrew Mason

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    This may help you: plot v as a function of t assuming a constant deceleration. It will be a straight line with slope [itex]a = \Delta v/\Delta t[/itex]. The area under the graph is the distance. You can see that it is:

    [tex]s = \frac{1}{2}\Delta v\Delta t[/tex]

  6. Jan 9, 2005 #5
    many thanks
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